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Proof step 2 from proof bmin-one-03 (previous step)

# s = 1 ⊦ s.butmin = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# s = 1, s.butmin.min ∈ s.butmin, s.butmin.max ∈ s.butmin ⊦ # s.butmin = 0, s.butmin = ∅

Rule argument

Additional information

Used simplifier rules: Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;% ⊦ ¬ a < a; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic